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Scores on a standardized test are normally distributed with a mean of 350 and a standard deviation of 25. Approximately what percent of students scored between 360 and 380?

Respuesta :

(360) = 10/25 = .4
z(380) = 30/25 = 1.2
P(350 < x < 380) = normalcdf(.4, 1.2)
JeanaShupp

Answer: 22.95 %

Step-by-step explanation:

Given : Scores on a standardized test are normally distributed with

Mean : [tex]\mu=350[/tex]

Standard deviation : [tex]\sigma =25[/tex]

The formula to find z-score:-

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x=360

[tex]z=\dfrac{360-350}{25}=0.4[/tex]

For x=380

[tex]z=\dfrac{380-350}{25}=1.2[/tex]

The p-value = [tex]P(360<X<380)[/tex]

[tex]=P(0.4<z<1.2)=P(1.2)-P(0.4)\\\\= 0.8849303-0.6554217=0.2295086\approx0.2295[/tex]

In percent , the percent of students scored between 360 and 380 =

[tex]0.2295\times100=22.95\%[/tex]

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